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ExampleReVISIOn 3

It is required for the shown industrial building to :-

4 ton ~Coil

t30 ton

4 .0

3.0CraneCrane12 to n 12 to n1 15.0

Bracing system for col. I

3.05.0

6.0 6.0

1. Design col. 1 as a rolled col.2. Design an end gable of height 9.2 m" 2.5 m wall, window 1m, and end girts each l.9 m " , take into consideration all cases ofwind on the column.Spacing between end gables 6 m

3. Design and draw a welded rafter splice used to join 2 parts of 6 m each to make the rafter.Rafter sec is LP.E 360

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Solution1- CoilLoads from crane girder on col. bracket

'LL S I

______________ 12/t~0-------1-2~/i~0---Rx

W dl = 0.2 t / mRn = .2 x 6 = 1 .2 ton

Dead Load

Live Load

Lateral Shock

PT = Rn + Ru ( 1 +I)=1 .2 + 16 x 1 .25 = 21 .2 tonR n = 12 + 12 x 2/6 = 16 ton Px =R;= 1 .6 tonRx = 1 .2 + 1 .2 x 2/6 = 1 .6 ton Pdl = 1.2ton

Braking force BB = ( 12 + 12 ) 1 7 = 3.42ton

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Straininl action on the coL:

Re\ISlOn 3

Straining actions in plane are coming from Pr and Px and causes normal force and moment in plane.As there is no Hz member in the vertical bracing at the level of the crane girder, the brakingforce will cause moment out of plane.

ColI

4ton~ t30 ton

B/2 3.42/2= 1.7 ton

26mt32

BMD~

M in plane

3.2 mt

M out plane

'32

NFD

32

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Choice of sec 3 - 3 :M, = =26 m t N= =9 ton My = 0 mt

Check: My=

Mx=26mt~

( Y )~ t~ ~

Sec 3 - 3

Assume f= = 1 .0 t/crrr' Sx = 26 x 100/ 1 .0 = =2600 cnr'Choose B.F.I.B 400 as sec is subjected to M, and My

Lbin = 2 x 8 = 16 mLbout = =8.0 mAm==Lbx / r x= 1600117.1 = 93.5< 180Aout=Lby/ry=800/7.41 = = 107.9< 180A . m a x = =107.9> 100F e = = 7500 1 (107.9)2 = 0.643 t/cm ''F c a = = 9/198 = = .045 t/cm '

Fbx = = Mx / s, = 2600 12 8 8 0 = =0.9 t / cm 2Luact = = 800 ern Lumax = = 20 bf / ~ Fy = = 20 x 30 1 ~2 .4 = 387 < 800Section is non com pact or slenderFJ = = - N 1A +M , / Sx = -.045 + 0.9 = .855F2 = = - N 1A - M , 1 Sx = = -.045 - 0.9 = =-0.945

D w / t w = = ( 40 - 4 x 2 .4 ) / 1 .3 5 = 2 2 .5 < (1 90 / ~ F y ) x (2 + 'II ) ) = = Ill. 49C / tf= 1 5/2 .4 = =6.2 5 < 10.9Sec is non com pacta = M smaU/ M big = = - 9/2 6 = -.346Fltb = = 800 Af C , 1 (Lu d) = = 800 x 30 x 2 .4 x l.42 / (800 x 40) = 2 .55 > 1 .4

Applying interaction equation:.07 + 1.0 x 0.9 /1.4 =0.71 < 1.2 " Case B " safe.

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Re\lSlOn .\

Check sec 1 - 1 N = 30 tonM, = 20 mt My = 3.2 mt Mx=20mt

~~i~C OL b i n = 2 x 8 = 16 mLbout = 8.0 mA m = Lbx 1 rx = = 1600 1 17.1 = 93.5 < 180A o u t = = L b y 1 r y = 800 1 7.41 = 107.9 < 180Amax= 107.9> 100Fe = = 75001 (107.9)2 = = .643 t/cnr'F c a = 30/198 = .151 t/cm2

My = 3.2 mt

Sec 1 - 1Fca/Fc==.235>.15

C r u x = 0.85 frame pennited to swayFca = .151Fex = 7500 1 " ! v x 2=7500 1 93.52 = .858

Al = 1.031Cmy = I frame prevented with lateral load and hinged baseFca=.15IFey = 7500 j)\ ../ = 7500 1107.9 2 = .643

Az = 1.3Fbx =M, 1 Sx =2000 12880 =0.7 t 1 cm2Fby = My 1 Sy = 320 1 721 = = 0.44 t 1 cnr'Fbcy = 1.4 t 1 cm2 "always Fbcy= 1.4 t 1 cm2"Luact = 800 em Lumax = 20 bf 1~ Fy = 20 x 30 1\f2.4 = = 387 < 800Section is non compact or slenderF J = - N 1A +M, 1 Sx = -0.15 + 0.7 = 0.55F 2 = = - N 1A - M, I Sx = -0.15 - 0.7 = -0.85

D w l t w = (40 - 4 x 2.4 ) 1 1.35 = 22.5 < (190 I ((~Fy) x (2 + I ' ) ) = 90.8C It[= IS 12.4 =6.25 < 10.9Sec is non compacta = = Msmall 1Mt,ig = 0

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RenslOn :;

Cs= 1.75 + 1.05 a +.3 a:::= 1.75Fltb = 800 Af Cb / ( Lu d ) = 800 x 30 x 2.4 x 1 .75 I ( 800 x 40 ) = 3.15 > 1.4Fbc x = 1.4 t I cm '

Applying interaction equation:.235 + 1.031 x.7 11.4 + 1.3 x 0.445/1.4 = 1.16 < 1.2 safe.

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ReVlSlOn 3

2 - End gable column.1- The acting loads:

h= 9.2 ma. Dead Load:

1. Own weight 50 kg/m'.11 . Weight of the steel sheets We= 6 kg/m '.lll. Weight of the end girts wg= 20 kg/m'.

Case of pressureb. Wind load:

1. Inthis case the wind load will be a main load so it will be a case A11 . Wwmd = (C, x K x q) X Sj Hz Bl"'Qdng

Ce= 0.8K = 1 .0 for h ~ 10mq= 70 kg/rrr' in Cairo

2- The straining actions: 'Wine! Loa.e! c : = [ : : >a. Wx= due to wind load only

Wx= Wwind= 0.8 x 1x 70 x 6 = 0.336 tim'il~1= Wx xh2 ?_~v = 0.336 x 9.2- / 8 = 3.55 mt

A 8

b. N= due to dead loads only v + X yN=wex(Sj xh)+o.w. x'h+wg x t number of girts j x S,N = 6 x (6 x 6.7) + 50 x 9.2 + 20 x (5) x 6 = 1.31 ton

Choice of sec:M, = 3.55 mt N= 1.31 tonAssume f= 1.2 t/cnr' Sx= 3.55 x 100/ 1.2 = 296 crrr'Choose I.P.E 240

Check:

Lbin = 2.5 m "wall height"

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Re\lSlOn ~

Lbout= 9.2 mA m = Lby 1 fy = 250 12.69 = 92.9 < 180Aout= Lbx 1 r, =920 19.97 =92.3 < 180A m . a x = 93 < 100Fe = 1.4 - 6.5 X 10-5 (93)2 = 0.83 tJcm2Fca= 1.31/39.1 = .033 tJcm2Fca/Fe=.043 1.4

ApplYing interaction equation:.043 + 1.0 x 1.09 11.4 =0.82 < 1 "case A" safe.

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Re\lSlOn)

Check on Case of Suctiona. Wind load:

ni. Inthis case the wind load will be a main load so it will be a case AIV. Wwind= (C, x K x q) X SI Hz Bra~

C;> 0.6K= 1.0 fo r h s10 mq = 70 kg/nr' in Cairo

3- The straining actions:a. Wx= due to wind load onlyWx=Wwind=0.6 x Ix 70 x 6 = 0.25 tim'

M = Wx X h2 = 0.25 X 9.22 /8 = 2.66 mtx 8

b. N= due to dead loads only y . + X yN= We X (SI x h) + o.w. x h + Wg x (number of girts) x S]N=6x(6x6.7)+50x9.2+20 x(5)x6= 1.31 ton

Choice of sec:Check on same section LP.E 240Check:

Lbin = 2.5 m "wall height"L o o m = 9.2 mA . m = Lby / fy = 250 / 2.69 = 92.9 < 180A o u t = Lbx / fx = 920/9.97 = 92.3 < 180" - m a x = 93 < 100Fe = 1.4 - 6.5 x 10-5 (93)2 = 0.83 t/cm''Fea = 1.31/39.1 = .033 t/cnr'F e a / F e = .043 < .15 Al = 1.0

Fbx = Mx / S x = 266 / 324 =0.82 t / cm2

l. ba_

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Luact = 920 em in this case as the compression flange IS the Inside flangeUse knee bracing at every end girts to brace the compression flangeLuact = 250 ern "wall height"Lumax = 20 bf I \i Fy = 20 x 12/ v l 2 A = 154.9 < 250Section is non compact or slenderFI = -NIA +M, / S x = = -.033 + 0.82 = 0.79F2 = -NIA-M, / Sx= -.033 - 0.82 = -0.85\fI=F1/F2 =-.92>-1D;/t, = ( 24 - 4 x 0.98 ) / 0.62 = 32.38 < (190 / ((~Fy) x (2 + 'P ) = 115.3C/t[=6/0.98=6.125< 14.8Sec is non compactCb= 1.13F1tb=800AfCb/(Lud)=800x 12 xO.98x 1.13/(250x22)= 1.7> 1.4Fbcx = = 1.4 t / cm2Applving interaction equation:.043 + 1.0 x 0.82 I 1.4 = 0.62

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Re\!SlOn 3

3 - Welded splicethe best way to make a welded splice is as the type II but using weld instead of bolts

As it was required to join parts of length 6m so use a splice at distance 6 m from col.At splice sec ;-

Qact = Y -WT . X of splice = 6 - 0.5 x 6 = 3 tonM m a x = S x X Fbcx = 1.536 x 904 = 1388 cmt

Lweld LweldI II 1I I II 0.8 h I II I II I

Mmax

iQactI, = 16270 cm4I w = 0.8 X 363 I 12 = 3110.4 ern"Mw = ( Iwl Ix) Mmax = 3110.4 116270 x 1388 = 266 cmtMf= Mmax - Mw = 1388 - 266 = 1122 cmt

A- Flange splice: " weld subjected to shear"

Lweld

1 MflangeJ

~~~5==1"_4I ~c IT = Me I hI = 1122 I 36 = 31.16 tonAssume S = 6 mmLweld act = T I (x Sx 0.2 Fu ) + 2 S = 3 1 . 16 I ( 6 x 0.6 x 0.72 ) + 2 x 0.6 = 13.2 = 15 em

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ReYlSlon 3

B- Web splice: " weld subjected to shear and torsion moment ..10cm

10.8 hI I " ,I JMwebI II I III I II 0.8 h I !I I I

J J

1 MwebIItQact

Assume S = 6 rnm , 1hzweld = 10 em . Ivl weld = 0.8 hr= 28 cmProperties of weld:A wvl :: 28 x 0.6 :: 16.8 crrr'A whz= 2 x 10 x 0.6 = 12 cnr'A wtot = 12+ 16.8 =28.8 cm2X = ( 10 x 0.6 x 2 x ( 10 I 2 + 0.6) + 28 x 0.6 x 0.3) 128.8 = 2.5 emI, = 0.6 X 283 112 + 10 x 0.6 x 2 x (28 12 + 0.6 12 i= 3551.48 ern"Iy= 0.6 X 103 I 12 + 2 x 10 x 0.6 x ( 5.6 - 2.5 )2 + 28 x 0.6 x ( 2.5 - 0.3 )2 = 296.632 em"Ip:: 3551.48 + 296.632:: 3848.11 em"

28 -

bMweb Mweb+Qact ( b)

x tQact tQactCheck on critical points:

M, = 266 + 3 ( 10.6 - 2.5 ) = 290.3 cmtEach side is subjected to M, I 2 =290.3 12= 145.1 cmt and Qactl 2 = 3 12 =1.5 ton

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RC\lSlOn 3

Point 1 :Rx= O+MtYj IIp = 0+145x(14)/3848.11=0.52t/cm2R, = Q a c t 1Awvl + M, XI IIp = 1.51 16.8 + 145 x 2.5 13848.11 = 0.18 t/cm2R =." j (R} + R/) = ,j (0.182 + 0.5i) = 0.55

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1. Design a typical secondary beam if the total dead load IS 400 kg/m (which includes the slab weightand the floor cover) and the live loads is 300 kg/rrr'. Choose an IPE section for this secondary beamand consider it to be simply supported on the main beam.

2. If the sections chosen for the main beams are IPEs 500, design the bolted connection between asecondary beam and a main beam using ordinary bolts 16 mm in diameter and 80x8 connectingangles (connection A in Figure 2).

3. Draw to scale I: 10 the above connection in different views.

Question No.3 (20% of the maximum credit)An area 12 x 40 Inwill be covered with the frames shown in Figure 3. The spacing between the framesis 5.0 m. End gables arc provided with columns along the two ends of the area as shown in the figure. Themain frames are composed of simple beams resting on pendulum columns. In this way, the system isunstable and can not carry any lateral loads.1. Suggest a suitable bracing system thatwould provide stability for the frames

against lateral loads and draw this system(scale 1 : 1 00) in different views. 6 m

2 . If the wind pressure in this area is 100kg/nr', calculate the forces in all thevertical bracing members due to windload.

3. Design suitable sections for the differentvertical bracing members.

I I I I I

6x2=12m

. _--. --- * ------"" - - - .._ . _ . _ . _-. - - - . . . . . . . . . . . . . . . . . _ _ _ _.. - - - .. :Figure 3